Given a vector space 𝑉 in ℝ^𝑛 and basis vectors {𝑒₁, …, 𝑒_𝑛} that spans 𝑉, one possible dual basis {𝜀¹, …, 𝜀^𝑛} that spans the dual space of 𝑉 can be defined as (where 𝛿 is the Kronecker delta function):

• ${𝜀}^i\cdot e_j={𝜀}^i(e_j)={𝛿}_j^i=\{\begin{array}{ll}1& ifi=j\\ 0& otherwise\end{array}$

In other words, we find a set of linear functionals {𝜀¹, …, 𝜀^𝑛} such that it “consumes” 𝑉‘s basis vectors {𝑒₁, …, 𝑒_𝑛} in the following way:

• ${𝜀}^i(e_j)=\{\begin{array}{ll}1& ifi=j\\ 0& otherwise\end{array}$

In other words, let:

• 𝐸 = [𝑒₁|𝑒₂|…|𝑒_𝑛] a matrix whose columns are the basis vectors {𝑒₁, 𝑒₂, …, 𝑒_𝑛}
• 𝐸ˆ = [𝜀¹|𝜀²|…|𝜀^𝑛] a matrix whose columns are the epsilon covectors {𝜀¹, 𝜀², …, 𝜀^𝑛}

The system of equations on the LEFT can be expressed as:

• ${\hat{E}}^{T}E=I$
• $\left[\begin{array}{cccc}{𝜀}_1^1& {𝜀}_2^1& \dots & {𝜀}_n^1\\ {𝜀}_1^2& {𝜀}_2^2& \dots & {𝜀}_n^2\\ ⋮& ⋮& \ddots & ⋮\\ {𝜀}_1^n& {𝜀}_2^n& \dots & {𝜀}_n^n\end{array}\right]\left[\begin{array}{cccc}{e}_{11}& e_{21}& \dots & e_{n1}\\ e_{12}& e_{22}& \dots & e_{n2}\\ ⋮& ⋮& \ddots & ⋮\\ e_{1n}& e_{2n}& \dots & e_{nn}\end{array}\right]=\left[\begin{array}{cccc}1& 0& \dots & 0\\ 0& 1& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 1\end{array}\right]$

TL;DR

1. Write your basis vectors as column vectors in a matrix
2. Invert that matrix
3. Your dual basis vectors are the row vectors of your inverted matrix

In other words, the dual basis which are the columns of 𝐸ˆ can be computed as:

• 𝐸ˆ = (𝐸^-1)^T